Question

need help solving this function y= x^2 -4 and also y=-3x^2-6x-5

Answer 1

are the functions equal to zero?

Answer 2

no i have to solve then graph them

Answer 3

i have tried them both but i am not sure they are correct

Answer 4

what are you solving for?

Answer 5

Since they both are in terms of y, set them equal to each other and then solve for x. This will be the x value at which the two functions intersect. Plug this x value back into either equation to find the point x, y at which the lines merge.

Answer 6

i need ot find the vertex if any the x intercept and y intercept

Answer 7

can you rephrase please

Answer 8

i need to find the vertex the xintercept and y intercept which would make them a parable

Answer 9

x-intercept is the point(s) where y = 0. y-intercept is the point(s) where x = 0. To find the x-intercept of \( y = x^2 - 4\), solve\[0=x^2-4\]\[x^2=4\]What values satisfy that equation? Hint: remember that a negative number times a negative number gives you a positive number.

Answer 10

it would be 2

Answer 11

or -2

Answer 12

Correct, those are the x-intercepts of y = x^2 - 4. Now can you find the y-intercept(s)? Proceed similarly, except this time you set x = 0 and find the value of y that makes it happen.

Answer 13

(it's easy!)

Answer 14

y would be 0,-4 but wouldn't that be the vertex

Answer 15

Yes, the y-intercept is at (0, -4). This is a parabola that is shifted down 4 units from the origin.

Answer 16

so then their is no vertex

Answer 17

Oh, there's always a vertex if it's a parabola! Can you tell me what the vertex of a parabola is? (not the value, just describe it)

Answer 18

the vertex of parabola is point where it crosses the axis

Answer 19

Um, no, not necessarily. Though many of the parabolas you've probably encountered might have their vertices there, it isn't generally true.

Answer 20

x marks the spot of the vertex of the parabola, even though it doesn't happen to fall on either axis

Answer 21

so they can be anyplace then

Answer 22

they show the highest or lowest point

Answer 23

If the parabola points up or down, yes, it's the bottom or the top. But you can have a parabola that is on its side, too! \(y^2=x\) would be an example.

Answer 24

So we better figure out the general way to find the vertex, eh? :-)

Answer 25

i would use the formula -b/2a

Answer 26

here their is no b so it would be -0/2(1)

Answer 27

That would be a good way. The vertex of y = x^2 -4 is at (0,-4). We get the 0 from the formula, and the -4 from plugging the 0 into the equation of the parabola.

Answer 28

(btw, "there is no b" is correct, "their" is a possessive pronoun, as in "that is their homework for the day")

Answer 29

yes that is what i did

Answer 30

Now how about that other parabola?

Answer 31

\[y=-3x^2-6x-5\]

Answer 32

We need to find x-intercept(s), y-intercept(s), and vertex. Shall we race? :-)

Answer 33

this one i found the vertex

Answer 34

which is -1,-2

Answer 35

and yintercept is 0,-5

Answer 36

then i solved with quadratic formula of -b plus or minus the square root of b^2 -4ac/2a

Answer 37

so what do you get for the x-intercepts?

Answer 38

(this might be regarded as a bit of a trick question)

Answer 39

when i did this i ran into a problem

Answer 40

okay...what were the solutions the quadratic formula gave you?

Answer 41

you cannot take the square root of -24 so then thier is no answer

Answer 42

Well, you can take the square root of -24. You just get what's called an imaginary number! As if the other kind are not imaginary, right? :-)

Answer 43

I take it you haven't encountered \( i = \sqrt{-1}\) yet?

Answer 44

no i have not

Answer 45

i was told if we have a negative in the square root we cannot do the problem then it is done

Answer 46

Okay, well, as it turns out, I'm reading a book called "An Imaginary Tale: The Story of \(\sqrt{-1}\)"! It seems that mathematicians of old (prior to 1600 or so) had a real fear of negative numbers, and would write their problems so that no negative numbers appeared.

Answer 47

sound like a good book

Answer 48

Then someone figured out that it was actually a pretty useful concept...

Anyhow, the 10 cent tour: if you let \(i^2 = -1\rightarrow i=\sqrt{-1}\) then you can factor your negative number under the square root sign like this:
\[\sqrt{-24} = \sqrt{-1*24} = \sqrt{24} * \sqrt{-1} = 2i\sqrt{6}\]

Answer 49

And there's a really cool result involving \(i, \pi, e, -1, 0\) (the big 5 numbers in math, you might say):
\[e^{i\pi} - 1 = 0\]
Who would think that those numbers from all over, so to speak, would come together in such an elegant way? What does the ratio of the circumference of a circle to its diameter have to do with the base of the natural logarithm, or the square root of -1? But there it is!

Answer 50

we were not taught to do this though..who wrote this bool

Answer 51

sorry book

Answer 52

Anyhow, back to our parabola. You don't get any answers that are real numbers, only answers that have \(i\) in them. That means there are no x-intercepts in this case (which is why I called it a trick question when I asked you what they are).

Answer 53

so i was right then no xinterecept

Answer 54

i am not crazy then

Answer 55

Here's a graph of our parabola:

Answer 56

to graph this how would i even do it

Answer 57

it would go down but not touching the xaxis

Answer 58

The book is by a guy named Paul Nahin. It's got a lot of complicated math in the first chapter, and more in the rest of the chapters, so it might be a difficult book to read, but the history parts and the descriptions of many different ways in which \(i\) comes into play is interesting.

Answer 59

Graph the way you graph any other function: make a table of x and y values and plot the points. The trick is figuring out which area of the infinite x-axis and y-axis is likely to be interesting!

Answer 60

could i use say 1 for x and plug it into the problem to add points

Answer 61

You can sort of assume at your level that probably the interesting area isn't going to be too far from the origin, but eventually you may not be able to make that assumption.

Yes, I would make a table of say x = -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 and then grind out the values of y for each x. Maybe do every other one at first. I always do 0, because that usually is pretty easy to evaluate :-)

Answer 62

thank you